WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today is the first in a series of lectures on conic sections.
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And we are going to start out with a review of two formulas that you will be applying later on, when we work with circles and hyperbolas and other conic sections.
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Starting out with the **midpoint formula**: this formula gives you the midpoint of line segments with endpoints (x₁,y₁) and (x₂,y₂).
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So, let's take an example, just to illustrate this: let's say you were asked to find the midpoint of a line segment with endpoints at (2,3) and (5,1).
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If these were given as the endpoints, and you were asked to find the midpoint, you could go ahead and apply this formula.
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So, to visualize this, let's draw out this line segment: this is at (2,3); that is one endpoint, and (5,1) is the other endpoint.
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I draw a line between these two; and I am looking for the midpoint, which is somewhere around here.
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Looking at this formula, it is actually somewhat intuitive, because what you are doing is finding
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the average of the x-values and the average of the y-values, which will give you the middle of each (the midpoint).
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So, x₁ + x₂ would give me 2 + 5 (the two x-values), divided by 2.
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That is going to give me the x-coordinate of the midpoint of this line segment.
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For the y-coordinate, I am going to add the two y's and divide by 2; so I will average those two y-values, which are 3 and 1.
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2 + 5 gives me 7/2; 3 + 1 is 4/2; I can simplify to (7/2,2); and I could rewrite this, even, as (3 1/2,2) to make it a little easier to visualize on the graph.
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This is the midpoint; it is going to be at 3 1/2 (that is going to be my x-coordinate); and 2 is going to be my y-coordinate.
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And that is the midpoint; this is a fairly straightforward formula; you will need to apply it in a little while, when we start working with circles.
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Also, to review the distance formula: if you recall, the distance formula is based on the Pythagorean theorem.
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And the distance formula tells us that we can find the distance between two sets of points
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if we take the square root of x₂ - x₁, squared, plus y₂ - y₁, squared.
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Now recall: if I am given two points; if I am asked to find the distance between a set of two points, (2,1) and (5,6),
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I can assign either one to be...I could assign this as (x₁,y₁), (x₂,y₂);
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or I could do it the other way around: I could say this is (x₂,y₂); this is (x₁,y₁).
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It doesn't matter, as long as you assign it and stick with that; you don't mix and match and say this is (x₁,y₂).
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Just assign either set; it doesn't matter which set you call which; just stay consistent with the order that you use the points in.
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OK, so let's look at what this would graph out to.
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(2,1) and (5,6) would be right about up there; so I have a line segment, and I am asked to find the distance.
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So, the distance is going to be equal to the square root of...I am going to call this (x₁,y₁) and this (x₂,y₂).
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So, x₂ is 5, minus x₁, which is 2; take that squared,
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and add it to y₂, which is 6, minus y₁, which is 1, squared.
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Therefore, distance equals 3², plus 6 minus 1 is 5².
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The distance equals the square root of 3², which is 9, plus 25.
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Therefore, distance equals √36; distance equals 6.
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The distance from this point to this point is equal to 6.
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And again, this is review from an earlier lecture, when we discussed the Pythagorean theorem in Algebra I
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and talked about how the distance formula comes from that.
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So, you can always go back and review that information; but this is the application of the distance formula.
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In the first example, we are asked to find the midpoint and distance of the segments with these endpoints (-9,-7) and (-3,-1).
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So, recall that the midpoint formula just involves finding the average of the x-coordinates
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of the two points, and the average of the y-coordinates of the two points.
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Applying these values to this set of equations gives me -9 + -3, divided by 2, and -7 + -1, divided by 2.
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Adding -9 and -3 gives me -12, divided by 2; and that is -8 divided by 2.
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This becomes -6 (-12/2 is -6), and then -8/2 is -4.
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So, that is the midpoint; that was the midpoint formula.
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Recall that the distance formula is (x₂ - x₁)² + (y₂ - y₁)².
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I am going to assign this as (x₁,y₁) and this as (x₂,y₂).
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Again, it doesn't matter; you can do it the other way around.
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Distance equals...I have x₂ as -3, minus -9, all of this squared; plus y₂, which is -1, minus -7, and that whole thing squared.
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This gives me -3; a negative and a negative is a positive, so plus 9, squared, plus -1...and a negative and a negative gives me + 7...squared.
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Therefore, distance equals the square root of...9 - 3 is 6, squared, and then 7 - 1 is also 6, squared.
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Distance equals √(36 + 36), or radic;72.
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But I can take this a step farther, and say, "OK, this is the square root of 36 times 2," and then simplify that,
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because this is a perfect square, to 6√2.
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So, I found the midpoint; it is (-6,-4); that is the midpoint of this segment with these endpoints.
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And the distance between these endpoints is 6√2.
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In Example 2, we are asked to find the midpoint and distance of the segment with these endpoints.
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We go about it as we usually do; but this time we are working with radicals,
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so we have to be really careful that we keep everything straight.
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Recall that the midpoint formula is the average of the x-coordinates, and then the average of the y-coordinates.
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Therefore, the midpoint equals x₁...the square root of 2 + 3, plus x₂, 2√2 - 4√3, divided by 2;
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for the y-coordinate of the midpoint, we are going to get √3 - 5, plus 4√3, plus 2√5.
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And then, we simplify this as much as we can.
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Here, I have two like radicals: they have the same radicand, so I can combine those to make this 3√2.
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I have a constant, and then this -4√3.
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So, this is the x-coordinate of the midpoint; the y-coordinate is √3 here and 4√3; those can be combined into 5√3;
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and then, let's see: this actually should have a radical over it, because that is a radical up there;
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I have -√5 + 2√5; that is going to leave me with just + √5, divided by 2.
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And the midpoint is given by this; it is a little bit messy-looking, but it is correct--you can't simplify, really, any farther.
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So, we are just going to leave it as it is.
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Now, we are also asked to find the distance; so I am going to go ahead and work that out here.
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We found the midpoint; we are working with these same endpoints, but we are finding the distance.
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Recall that the distance formula is the square root of (x₂ - x₁)² + (x₂ - y₁)².
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So, I am going to let this be (x₁,y₁), (x₂,y₂).
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I could have done it the other way around; it doesn't matter, as long as you are consistent.
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Therefore, the distance is going to be equal to y₂, which is 2√2, minus 4√3;
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and I am going to take that, and from that I am going to subtract y₁, which is √2 + 3.
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So, this covers my (x₂ - x₁)².
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I am going to add that to (y₂ - y₁): I go over here, and I have y₂; that is 4√3 + 2√5.
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And I am going to subtract y₁ from that, which is over here; and that is √3 - √5; and this whole thing is also going to be squared.
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Let's apply these negative signs to everything inside the parentheses, so we can start doing some combining.
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This is going to give me 2√2 - 4√3; negative...that is going to give me -√2;
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apply the negative to that 3--it is going to give me -3; all of this is going to be squared.
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Plus 4√3, plus 2√5...apply the negative to each term inside the parentheses to get -√3;
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and this negative, and then the negative √5, gives me a positive √5; and this whole thing is squared.
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All right, so let's see if I can do some combining to simplify a little bit before I start squaring everything,
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because that is going to be the most difficult step.
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All right, this gives me 2√2, and this is minus √2; so I can combine these two to get √2.
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-4√3, minus 3; this whole thing squared, plus 4√3 - √3--that simplifies to 3√3.
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2√5 + √5 gives me 3√5; squared.
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Now, I need to just square everything, and then combine it.
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There is no easy way around this first one; what I am going to do is multiply √2 times itself, times this, times this,
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and then go on with the second, and finally the third, term.
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So, this √2 times √2 is simply going to give me 2; √2 times -4√3 is going to give me -4... 2 times 3 is √6.
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Then, √2 times -3 is going to give me -3√2.
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OK, now I take -4√3 and multiply it by this first term, √2, to get -4; 3 times 2 is 6.
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When I multiply this times itself, I am going to get -4 times -4; that is going to give me 16,
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times √3 times √3 is going to give me 3.
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Then, I multiply this times -3 to get -4 times -3 is 12√3.
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Finally, multiplying -3 times √2 gives me -3√2; -3 times -4 is 12√3; -3 times -3 is 12√3.
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And then, -3 times -3 is 9; all of this is the trinomial squared; now let's square the binomial.
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3√3 times 3√3 is going to be...3 times 3 is 9; and then √3 times √3 is just going to be 3.
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So, looking up here, just to make this a little clearer: 3√3 + 3√5, times itself (squared)...
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I am going to do my first terms, and I get this; then I do the outer terms, plus the inner terms,
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which would just be this times this times 2; so that is going to give me 2 times 3 times 3, which is 9,
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times 3 times 5, with a radical over it; so that is this.
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So, what I did is took...really, it is just outer plus inner, which is going to give me the outer, which is 9√15, plus the inner, which also 9√15.
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And this is going to end up giving me 18√15, which is the same as what I have here.
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And then, finally, the last terms are going to give me 3 times 3 is 9, and then √5 times √5 is just going to give me 5.
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OK, now combining what we can in order to just make this a bit simpler: I have a constant here;
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I am just going to cluster all of my constants together in the beginning; that is 2; 16 times 3 is 48, so that is 48;
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what other constants do I have?--I have a 9, and then I have 27, and I have 45.
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All right, so those are my constants.
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Now, for terms with a √6 in them, I have 2 of those: -4√6 and -4√6 is going to give me -8√6.
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So, I took care of the constants and the terms with a radicand of 6.
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Next, let's look at terms with a radicand of 2: -3√2, and I have -3√2 here; and that is it.
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I add those two together, and I am going to get -6√2.
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So, these are all taken care of: √2, √6, constant: now I have √3.
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12√3--do I have any other terms like that?--yes: 12√3, so I have 2 of those; and that is going to be...12 and 12 is 24√3.
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Let's see what else I have left: I took care of those, those...that just leaves me with...2 times 9 is 18 times √15.
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Therefore, the distance equals...putting this all together, if you added these up, you will get 131 - 8√6 - 6√2 + 24 √3 + 18 √15.
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So, this is the distance; and this was really a lot of practice just working with multiplying and adding and subtracting radicals.
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We found the midpoint in the previous slide; and here is the distance for the segment with these endpoints, applying the distance formula.
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Example 3: Triangle XYZ has vertices X (4,9), Y (8,-9), and Z (-5,2).
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Find the length of the median from X to line YZ.
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Definitely, a sketch would help us to solve this; so let's sketch this triangle.
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And it has...we will call this X at (4,9); at (8,-9), we are going to have Y; and then, Z is going to be over here at (-5,2).
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The median is going to be a line going here from here right to this midpoint.
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So, what we are asked to find is the length: we can use the distance formula to find the length.
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But in order to use the distance formula, I need this endpoint and this endpoint.
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But I don't have this endpoint; however, I can find it, because if you look at this, this is going to draw out right to here, right in the middle.
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So, this is the midpoint; if I find the midpoint of YZ, then I have the endpoint of this median.
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Using the midpoint formula: (x₁ + x₂)/2, and then for the y-coordinate, (y₁ + y₂)/2.
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So, I am looking for the midpoint of YZ: that is going to give me...
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for Y, the x-value is 8; for Z, it is -5; divided by 2; for Y, the y-value is -9; for Z, the y-value is 2; divided by 2.
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This is going to give me a midpoint of 8 - 5; that is 3/2; -9 + 2 is -7/2.
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This is the midpoint, which is (3/2, -7/2); now I can use my distance formula,
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distance equals √((x₂ - x ₁)² + (y₂ - y₁)²).
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I can use that distance formula to find this.
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And I am going to call (4,9)...I need the distance from x to this point m...I am going to call this
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(x₁,y₁), and this (x₂,y₂), applying my distance formula.
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So first, x₂, which is 3/2, minus x₁, which is 4--I am going to rewrite that as 8/2 to make my subtraction easier.
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But this is just 4; squared; plus y₂, which is -7/2, minus 9, which I am going to rewrite as 18/2.
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Again, I am just moving on to the next step of finding the common denominator.
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But this is my (x₁,y₁), (x₂,y₂); that is where these came from.
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Therefore, the distance equals...3/2 - 8/2 is -5/2; and we are going to square that;
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plus -18 and -7 combines to -25/2, squared.
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Therefore, the distance equals the square root of...this is 25 (5 squared), over 4, plus...25 squared is actually 625, divided by 4,
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which equals the square root of 625 and 25 is 650, and they have the common denominator of 4.
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And you could leave it like this, or take it a step farther and write this as 650 times 1/4, which equals 1/2.
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And you could even look and see if there are any other perfect squares that you could factor out.
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But this does give us the length of the median; in order to find the length of the median, we had to find this other endpoint,
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which conveniently was the midpoint of YZ, so we found the midpoint of YZ and used that as the other endpoint.
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Then I had X as an endpoint and the midpoint as the endpoint.
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Plug that into the distance formula to get the length.
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Working with triangles again: find the perimeter and area of the triangle with vertices at (1,-4), (-1,-2), and (6,1).
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So, to help visualize this, we are going to draw it out on the coordinate plane.
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(1,-4) is my first point; (-1,-2) is my second point; and then 6 is going to be about here; this is 6; 1 will be right here.
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Let's see what this triangle looks like, and if the graph can help me.
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Now, I am asked to find the perimeter and area: if I want to find the area of a right triangle,
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I know that it is going to be 1/2 the base times height.
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The problem is that I don't know if I am working with a right triangle; I may be; but this definitely does not look like a right angle.
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This may be a right angle, but I am not sure.
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In order to determine if it is a right angle (let's call this A, B, and C)--is this a right angle?--
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well, if it is a right angle, that means that this AC and BC are going to be perpendicular.
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And recall that perpendicular lines...the product of their slope equals -1.
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So, what I am going to do is go ahead and find the slope of these two.
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And I am going to determine what this slope is of AC; what BC is; find their product; and if they are a right angle, then I can proceed.
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So, this point right here is (-1,-2); this point is (1,-4); and this point is (6,1).
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Slope is just change in y, divided by change in x.
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So, let's find the slope of AC: that is going to be the change in y (-2 - -4), divided by change in x (which is -1 - 1).
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So, this is going to be -2 plus 4, divided by -2; this is going to give me 4 divided by -2, which is -2.
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I just double-check this: -2 minus -4...actually, correction: 4 minus 2 is going to give me, of course, 2; and the slope, therefore, will be -1.
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OK, now I also need to find, over here (to keep this separate), the slope of BC.
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So, change in y: y is 1 - -4, over change in x: x is 6 - 1; this is going to be equal to 1 + 4, divided by 6 - 1 is 5, which equals 5/5.
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So, the slope of BC equals 1.
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Now, this means that, if I have the slope of BC, times the slope of AC, equals 1 times -1, or -1.
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This is a right angle; I have a right triangle; I can use my formula, 1/2 the base times the height.
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What I need to do next is find the length of these sides, using the distance formula.
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So, let's start out by finding the length of side AC.
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Recall your distance formula: for the distance formula, I am going to take x₂, side AC,
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and I am going to call...it doesn't matter which one...I am going to call this, for this first one, (x₁,y₁), (x₂,y₂).
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x₂ is -1, minus x₁, which is 1, squared, plus y₂, which is going to be -2, minus y₁, which is -4.
00:28:24.700 --> 00:28:43.300
So, the length of AC equals...-1 and -1 is -2, squared; plus -2...and a negative and a negative is a positive, so -2 + 4 is going to give me 2, squared.
00:28:43.300 --> 00:28:53.100
Therefore, the length of AC is going to equal the square root of 4 plus 4, which equals the square root of 8.
00:28:53.100 --> 00:28:54.900
So, that is the length of AC.
00:28:54.900 --> 00:28:59.600
Now, I need to find the length of another side to find the area.
00:28:59.600 --> 00:29:08.300
I found AC; let's go for BC next--the length of BC.
00:29:08.300 --> 00:29:13.100
I am going to make this (x₁,y₁),(x₂,y₂).
00:29:13.100 --> 00:29:21.400
So, I am starting out with my x₂, which is 6, minus x₁, which is 1, squared,
00:29:21.400 --> 00:29:33.200
plus x₂, which is 1, minus -4 (that is y₂ - y₁), squared,
00:29:33.200 --> 00:29:42.700
equals the square root of...6 minus 1 is 5, squared; 1 minus -4...this becomes 1 + 4, so that is 5, squared;
00:29:42.700 --> 00:29:49.600
So, this equals the square root of 25 + 25, or the square root of 50.
00:29:49.600 --> 00:30:04.400
All right, so now I have two sides: I know that this side, AC, has a length of √8; and I know that BC is √50.
00:30:04.400 --> 00:30:27.800
So, the area equals 1/2 the base times the height; so the area equals 1/2 (√8)(√50), which equals 1/2√400.
00:30:27.800 --> 00:30:45.600
All right, and you could go on and then simplify this, because this would give you the perfect square of 20 times 20,
00:30:45.600 --> 00:30:56.600
because 20² would give you 400; so then I could make this 1/2(20) is 10.
00:30:56.600 --> 00:30:59.400
All right, now we still need to find the perimeter.
00:30:59.400 --> 00:31:04.500
In order to find the perimeter, I need this third side; and I can use the Pythagorean theorem, because this is the hypotenuse.
00:31:04.500 --> 00:31:08.800
And I know that a² + b² = c².
00:31:08.800 --> 00:31:19.300
So, a² is...one side is the square root of 8, squared, plus b², the square root of 50 squared, equals c².
00:31:19.300 --> 00:31:29.800
Well, the square root of 8 squared is 8, plus 50 (because the square root of 50 squared is going to be 50), equals c².
00:31:29.800 --> 00:31:35.400
So, 58 = c²; therefore, c equals √58.
00:31:35.400 --> 00:31:46.900
All right, so I found the area right here; now, the perimeter.
00:31:46.900 --> 00:31:59.900
The perimeter equals the sum of the three sides: so that is √8 + √50 + √58.
00:31:59.900 --> 00:32:02.500
And you could go on and do a little simplification.
00:32:02.500 --> 00:32:06.800
You could pull the perfect squares out of here.
00:32:06.800 --> 00:32:15.000
So, I could go a little farther with this and say, "OK, 8 is equal to 4 minus 2; 4 is a perfect square, so this is 2√2."
00:32:15.000 --> 00:32:22.600
50 is the square root of 25 times 2, so that is going to give me 5√2, plus √58.
00:32:22.600 --> 00:32:26.900
But you can't combine any of these, because they are not like radicals.
00:32:26.900 --> 00:32:32.800
So, in this example, we applied the distance formula, as well as the Pythagorean theorem and some geometry,
00:32:32.800 --> 00:32:37.300
to find the area and the perimeter of this triangle.
00:32:37.300 --> 00:32:42.000
Thanks for visiting Educator.com; and that concludes this lesson on the midpoint and distance formulas.